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Chapter 4: Problem 41

Multiple Choice A cylindrical rubber cord is stretched at a constant rate of 2 \(\mathrm{cm}\) per second. Assuming its volume does no change, how fast is its radius shrinking when its length is 100 \(\mathrm{c}\) and its radius is 1 \(\mathrm{cm} ?\) $$\begin{array}{ll}{\text { (A) } 0 \mathrm{cm} / \mathrm{sec}} & {\text { (B) } 0.01 \mathrm{cm} / \mathrm{sec}} 67 {\text{ (C) } 0.02 \mathrm{cm} / \mathrm{sec}}$\\\ {\text { (D) } 2 \mathrm{cm} / \mathrm{sec}} & {\text { (E) } 3.979 \mathrm{cm} / \mathrm{sec}}\end{array}

Short Answer

Expert verified
The radius is shrinking at a rate of 0.01 cm/sec, so option (B) is the correct answer.

Step by step solution

01

Differentiate the Volume Formula

We know the formula for the volume of a cylinder is \(V = 蟺r^2h\). We take the derivative of both sides of this equation with respect to t, the time, using the chain rule. This yields \(\frac{dV}{dt} = 蟺 (2r\frac{dr}{dt}h + r^2\frac{dh}{dt})\).
02

Substitute Given Values

As per the given problem, the rate of change of height, \(\frac{dh}{dt}\) is 2 cm/sec, the radius, r is 1 cm and the length when the rate is desired, h is 100 cm. Also, it is known that volume does not change, so \(\frac{dV}{dt} = 0\). Substitute these values into the derived equation to solve for \(\frac{dr}{dt}\), the rate of change of the radius. We get \(0 = 蟺 (2*1*\frac{dr}{dt}*100 + 1^2*2)\).
03

Solve for The Rate of Change of the Radius

Solving the equation from Step 2 for \(\frac{dr}{dt}\) provides the rate at which the radius is changing. Simplifying the equation gives: \(\frac{dr}{dt} = -\frac{1}{100}\). This shows that the radius is shrinking at 0.01 cm/second (since the rate is negative).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Volume
Understanding the volume of a cylinder is fundamental in solving real-world problems involving cylindrical shapes. The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \], where \( V \) represents the volume, \( r \) is the radius of the cylinder's base, and \( h \) is the height of the cylinder.

When working with problems where the volume remains constant even though other dimensions change, comprehending this formula is crucial, because it allows us to relate the changing dimensions to one another. For instance, if either the radius or the height changes, and we wish to maintain the same volume, the other dimension must adjust accordingly.
Chain Rule Differentiation
The chain rule is a significant differentiation technique in calculus that allows us to calculate the derivative of a function composed of other functions. In simple terms, the chain rule helps us find the rate at which one thing is changing with respect to another when the two are related in some way.

For example, if we have a volume \( V \) that depends on the radius \( r \) and the height \( h \), and both radius and height change over time, the chain rule allows us to express the rate of change of the volume \( \frac{dV}{dt} \) in terms of the rates of change of radius \( \frac{dr}{dt} \) and height \( \frac{dh}{dt} \). Applying the chain rule would yield an expression that relates all these rates together, which is especially useful for solving related rates problems in calculus.
Rate of Radius Change
The rate of radius change refers to how quickly the radius of an object is expanding or shrinking over time. In the context of a cylindrical shape, when volume is conserved and height is increasing, the radius must necessarily be decreasing to compensate.

In our exercise, the cylinder is stretched, meaning the height increases. To find how fast the radius is shrinking at a specific moment, we use the formula derived from differentiating the volume of the cylinder with respect to time while keeping the volume constant. We solve for \( \frac{dr}{dt} \), which represents the rate of radius change, to understand how the radius alters as the height changes.
Volume Conservation
Volume conservation is a principle that states that in the absence of external forces, the volume of an object remains constant despite changes in its shape. In our related rates problem, we apply this principle by setting the rate of volume change, \( \frac{dV}{dt} \), to zero.

This assumption leads us to establish a relationship between the rate of change of the radius and the rate of change of the height of our cylinder. To maintain a constant volume, if one dimension of the cylinder increases, the other must decrease at a rate that keeps the overall volume unchanged. This concept is key in solving many physics and engineering problems where volume conservation is an important constraint.

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Most popular questions from this chapter

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